PLEASE CHECK - are my titration calculations correct? (1 Viewer)

[Lordsion]

Forever in debt if you answer this!!!!


Titration of NaOH with Vinegar (acetic acid). Vinegar is in the burette, NaOH in the flask.

I used NaOH with concentration 0.10 mol -1 and i used 25 ml of it.
My Titre of vinegar used to reach equivalence point of NaOH was 21 ml.

The vinegar was DILUTED 1 in 5.

I HAVE TO FIND (A) AND (B)

(a) molarity of diluted vinegar solution (which is 1 in 5 - i did 50 ml vinegar with 200 ml water = 250 ml)

(b) Molarity of original vinegar solution (undiluted 50ml)

MY SOLUTIONS ====== ARE THEY RIGHT METHOD?

(a) C1 X V 1 = C2 X V2

0.10 (concentration of NaOH) X 25 (ml of NaOH) = C2 (? concentration of diluted vinegar) X 21 (titre of vinegar used)

0.10 X 25 / 21 = C2

0.1190 ..... mol -1


(b) since i diluted the vinegar 4 times its volume .....

0.1190 .... X 4 (or is it 5? ) = 0.476 (this is the concentration of undiluted vinegar)



is this right ?

Also, is 0.10 Mol -1 the same as 0.100 mol -1 ???

 

[dan.121212]

i believe first part is correct, but second part - times by five, not four (i believe)

also, there is no such thing as "0.10 Mol -1" or "0.100 mol -1", - there IS such a thing as: 0.10 mol, 0.100 mol, 0.100 mol. L^-1, 0.10 mol.L^-1

the first two are number of moles, the latter two are measures of concentration in mol per litre

as for answering your Q - i answered it already on your 'Titration - are my steps right?' forum

 

[undalay]

Also, is 0.10 Mol -1 the same as 0.100 mol -1 ???

I assume you are refering to moles per litre.
In which case the first number has less significant figures then the second.
This may or may not alter your final result, however generally in HSC it doesn't matter that much.

 

[Lordsion]

i believe first part is correct, but second part - times by five, not four (i believe)

also, there is no such thing as "0.10 Mol -1" or "0.100 mol -1", - there IS such a thing as: 0.10 mol, 0.100 mol, 0.100 mol. L^-1, 0.10 mol.L^-1

the first two are number of moles, the latter two are measures of concentration in mol per litre

as for answering your Q - i answered it already on your 'Titration - are my steps right?' forum

THANK YOU SO MUCH - i have an assesment tom/ow and i am going through the questions making sure i have them right.

One more question though:

"How would the concentration of the NaOH solution in this experiment have been determined?"

I am gessing this:

The lab assistant would determine the amount of grams of NaOH needed to make up the concentration of 0.1 mol L^-1 by using this formula:

m = M x Mr x V. (Mass of substance = concentration of NaOH X molecular weight of NaOH X volume wanted ... say 250 ml) ?

I really need help with this one!

 

[Lordsion]

And also - that calculation in my initial post is worth 10 marks !!

Will i get ten makrs for those calculations - like, they dont seem like enough to get 10 marks.

 

[dan.121212]

DONT PANIC - ull be fine

 

[Lordsion]

DONT PANIC - ull be fine

But what do you think - about how it (concentration of NaOH) is determined?

I forget if any other rules are involved.

 

[Lordsion]

DONT PANIC - ull be fine

Lol - but it took me an hour just to figure out how to do those calculations

 

[WantToDoBetter]

But what do you think - about how it (concentration of NaOH) is determined?

I forget if any other rules are involved.

Since your test was the next day it's already done but ill let you know anyway.

The concentration of the solution would either be tested by titrating it with a primary standard solution.

Or if it was made to be the primary standard solution (and it wouldve been checked with a titration anyway) the lab assistan would have done the maths (can't remember) then weighed the sample to the appropriate weight, added it to the appropriate volume of water and shakin' and stired.

I think that's your answer.

Cheers.

 

[imnotkarl]

Forever in debt if you answer this!!!!


Titration of NaOH with Vinegar (acetic acid). Vinegar is in the burette, NaOH in the flask.

I used NaOH with concentration 0.10 mol -1 and i used 25 ml of it.
My Titre of vinegar used to reach equivalence point of NaOH was 21 ml.

The vinegar was DILUTED 1 in 5.

I HAVE TO FIND (A) AND (B)

(a) molarity of diluted vinegar solution (which is 1 in 5 - i did 50 ml vinegar with 200 ml water = 250 ml)

(b) Molarity of original vinegar solution (undiluted 50ml)

MY SOLUTIONS ====== ARE THEY RIGHT METHOD?

(a) C1 X V 1 = C2 X V2

Ok Hold up, this is not tiratraton, this is dilution.
C1V1=C2V2 is ONLY for dilution.

It will give you the answer to your question but is not the correct method to acheive a band 6 response.

Follow these steps for the correct TITRATION procedure:

Step 1: Write a Balanced Chemical Equation for the Reaction (of acid+base...B+A=S+W+CD)

Step 2: Calculate the moles of the known solution (n=m/M or n=C/V)

Step 3: Use the mole ratio to work out the moles of the unknown solution in the flask.

Step 4: Determine the concentration of the unknown solution using C=n/V.

Done presto!

Using the Dilution equation can only be used for the primamry standard but cannot yeild accurate results for the secondry standard.

In questions that give solutions with grams of acids diluted, it does not work.

cheers,
pat

 

[Lucid Scintilla]

Ok Hold up, this is not tiratraton, this is dilution.

Ok; hold up; this is titration and dilution.

Thread-starter has a question where volumes of an acid and a base reacting, with one's concentration known and the other to be found.

C1V1=C2V2 is ONLY for dilution.

Depends actually; what do subscripts '1' and '2' stand for? If you let '1' and '2' equal 'initial' and 'final', then it's dilution; if it's 'a' and 'b' (for 'acid' and 'base'), then it could be titration.

Will i get ten makrs for those calculations - like, they dont seem like enough to get 10 marks.

No, no you won't.

It will give you the answer to your question but is not the correct method to acheive a band 6 response.

'Band-6 response' tends to apply to extended-response questions.[/quote]

DONT PANIC - ull be fine

Generic advice is needless. Console yourself with it when you're panicking.

OP, your calculations for the reacting vinegar are correct. Your calc.'s for the initial concentration of the vinegar, prior to dilution, it would be multiplied by five.
(c_iv_i = c_fv_f, where c's are concentrations, and v's are (overall) volumes.)

Hey, good job with answering the OP's questions, people.